If the function f(x)=≤ft\{{array}{cll} a x- b x/ c x- b x & , { if } & x ≠ 0 \\ -1 & , { if } & x=0{array}. is continuous at x=0, then a², b², c² are in

Correct Answer is : Arithmetic progression

Arithmetico-Geometric progression

Correct Answer is : Arithmetic progression

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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 116 of 150

Marks: +1, -0

If the function

f(x)=\left\{\begin{array}{cll}\frac{\cos a x-\cos b x}{\cos c x-\cos b x} & , \text { if } & x \neq 0 \\ -1 & , \text { if } & x=0\end{array}\right.

x=0

a^{2}, b^{2}, c^{2}

are in

Solution:

Concept:

For continuity at

x = 0

, the limit of

f(x)

x \to 0

must equal

f(0) = -1

Explanation:

Since

f(x)

is continuous at

x = 0

, we have:

\lim_{x \to 0} f(x) = f(0) = -1

So,

\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos bx} = -1

Using the identity

\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)

, we get:

\lim_{x \to 0} \frac{-2 \sin\left(\frac{a+b}{2}\right)x \sin\left(\frac{a-b}{2}\right)x}{-2 \sin\left(\frac{c+b}{2}\right)x \sin\left(\frac{c-b}{2}\right)x} = -1

Since

\lim_{x \to 0} \frac{\sin kx}{kx} = 1

, the limit simplifies to:

\frac{\left(\frac{a+b}{2}\right)\left(\frac{a-b}{2}\right)}{\left(\frac{c+b}{2}\right)\left(\frac{c-b}{2}\right)} = -1

\Rightarrow \frac{a^{2} - b^{2}}{c^{2} - b^{2}} = -1

\Rightarrow a^{2} - b^{2} = b^{2} - c^{2}

\Rightarrow a^{2} + c^{2} = 2b^{2}

This means

a^{2}, b^{2}, c^{2}

are in arithmetic progression.

Answer:

B.

Arithmetic progression

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